Must have counted wrong. Counted again and you are right.
Great problems though. I cannot figure out how to conclude it is the solution you got. Do you do it by induction? I think I could probably get the answer by induction, but haven’t bothered trying.
Take the kth car. It is at the start of a cluster if it is the slowest of the first k cars. The kth car is therefore at the start of a cluster with probability 1/k. The expected number of clusters is the sum over all cars of the probability that that car is in the front of a cluster.
Must have counted wrong. Counted again and you are right.
Great problems though. I cannot figure out how to conclude it is the solution you got. Do you do it by induction? I think I could probably get the answer by induction, but haven’t bothered trying.
Take the kth car. It is at the start of a cluster if it is the slowest of the first k cars. The kth car is therefore at the start of a cluster with probability 1/k. The expected number of clusters is the sum over all cars of the probability that that car is in the front of a cluster.
Hurray for the linearity of expected value!